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IHM Ranchi admissions are completely based on merit. The institute offers admission only on the basis of valid entrance scores. In order to secure admission to the institute, candidates must appear for NCHMCT JEE and acquire the specified eligibility criteria. Candidates must follow the step-by-step process to secure admission to the institute.

Institute of Hotel Management Ranchi offers admissions to UG, certificate, and Diploma in the stream of Hotel Management. IHM Ranchi Admissions 2025 for all courses, are currently open for 2025. Interested candidates can apply to the desired course through its official website in online mode. IHM Ranchi admissions to all courses are merit-based.

Institute of Hotel Management offers admission based on the merit acquired in the last qualifying exam, along with valid entrance scores. In order to secure admission to IHM Ranchi, candidates must fulfil the course-specific eligibility criteria. Candidates can apply to the desired course online on the institute's website.

It's hard to tell, but this might be a query about the placement records for B.Tech in Plastic Engineering at CIPET Raipur. Unfortunately, without more information or context, a definitive answer cannot be provided.

No, only oscillations with a linear restoring force and sinusoidal motion are SHM. Non-linear oscillations, such as large pendulum swings, deviate from SHM.

Oscillations are larger, slower to-and-fro motions. Vibrations, on the other hand, are smaller, faster oscillations. Vibrations often occur in machines, strings, and structural components.

No, periodic motion isn't always oscillatory. It may not involve back-and-forth movement about an equilibrium point. Some known examples include circular motion in satellites or planets. 

Future Institute of Engineering and Management offers BTech courses in several specialisations. Moreover, the institute also offers lateral entry admission to some of the offered BTech courses. The names of some of the offered BTech specialsiations are as follows:

• Artificial Intelligence & Machine Learning
• Big Data & Analytics
• Civil Engineering
• Computer Science Engineering
• Cyber Security
• Electrical Engineering
• Electronics & Communication Engineering, etc. 

The primary eligibility requirement for admission to BTech LE courses at FIEM is that applicants must complete their BSc degree or Diploma in the relevant field with decent scores. Moreover, for the admission process, the institute accepts JELET scores. West Bengal Joint Entrance Lateral Engineering Test (WB JELET) is conducted annually by the West Bengal Joint Entrance Examinations Board (WBJEEB) to provide eligible candidates admission into the second year (third semester) of Bachelor's Degree courses in Engineering. 

The total fee structure for the BTech course might include multiple components, such as a tuition fee, a one-time fee, etc. Some of the fee components are required to be paid annually, semesterly or one-time. Check out the table below to know the exact fee structure:

$Mn{O}_4^{-2}\to A+B$  

Oxidation state of Mn in B < A 

$Mn{O}_4^{-2}+H^{+}\to Mn{O}_4^{-}+Mn{O}_2$  

B is MnO₂

$\therefore$  Oxidation state of Mn = +4

${\therefore }_{25}M{n}^{+4}=1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^{0}3d^3$  

  $\therefore$ unpaired electron = 3

Spin only magnetic moment  $\left(\mu \right)$  

$=\sqrt[]{n\left(n+2\right)}=\sqrt[]{3\left(3+2\right)}=\sqrt[]{15}$  

$=3.87\approx 4$  

$C\mathcal{l}{F}_3$ ->T-shaped (sp³d)

IF₇ ->Pentagonal bipyramidal (sp³d³)

BrF₅ -> Square pyramidal (sp³d²)

BrF₃ ->T-Shaped (sp³d)

I₂Cl₆ -> Triangular bipyramidal (sp³d)

$I{F}_5\to$  Square Pyramidal (sp³d²)

ClF -> (sp³)

ClF₅ -> Square Pyramidal (sp³d²)

Br₅, IF₅ & ClF₅ ® Square Pyramidal

t_1/2 = 0.301 min

t = 2 min

$K=\frac{2.303}{t}\mathcal{l}og\left(\frac{Co}{Ct}\right)$  

$\frac{0.693}{0.301}=\frac{2.303}{2}\mathcal{l}og\left(\frac{Co}{Ct}\right)$  

$\frac{2.303\times 0.301}{0.301}=\frac{2.303}{2}\mathcal{l}og\left(\frac{Co}{Ct}\right)$  

$\therefore 2=\mathcal{l}og\left(\frac{Co}{Ct}\right)$  

$\frac{Co}{Ct}=10^2=100$  

              Ans. 100

Ka for C₃H₇COOH = 2 × 10^-5

$pKa=-\mathcal{l}og\left(2\times 10^{-5}\right)=5-\mathcal{l}og2$  

=5 – 0.3 = 4.7

pH of 0.2 (M) solution = 

$pH=\frac{pKa-\mathcal{l}ogC}{2}$  

$=\frac{1}{2}\left(4.7\right)-\frac{1}{2}\mathcal{l}og\left(0.2\right)$  

$pH=27\times 10^{-1}$     

Ans 27